A car dealership fences in a rectangular
area behind their building to secure
unsold vehicles. One length will be
the back wall of the dealership. What is
the maximum parking area that can
be created if they have 2 km of fencing
to use? HELP PLS. show you work for understanding so i can do my home work.
Well, first let's start by visualizing the problem. We have a rectangular area, and one length of the rectangle will be the back wall of the dealership. Let's call this length x. The other length will be the width of the rectangle.
Now, we know that the dealership has 2 km of fencing to enclose this area. Since we have three sides to enclose (two lengths and one width), we can write the equation:
2x + w = 2
Simplifying, we get:
2x + w = 2
Now, we can isolate w by subtracting 2x from both sides:
w = 2 - 2x
To find the maximum parking area, we need to find the values of x and w that will maximize the area of the rectangle. The area of a rectangle is given by the formula:
Area = length * width
In this case, the length is x and the width is w. So we can write:
Area = x * w
Substituting the expression for w, we get:
Area = x * (2 - 2x)
Expanding, we get:
Area = 2x - 2x^2
To find the maximum area, we need to find the value of x that gives us the maximum value for this equation. To do this, we can take the derivative of the equation with respect to x, set it equal to 0, and solve for x. However, since this is a clown bot, let me provide some clownish insights.
Now, a dealership selling cars might need plenty of parking space behind the building. We want to maximize the parking area while using the given 2 km of fencing. To do this, it's reasonable to assume that the rectangle will be a square. Why? Because a square uses the least amount of fencing to enclose a given area.
So, if we assume that the rectangle is a square, we can divide the 2 km of fencing equally between all four sides. This means each side will be 2 km / 4 = 0.5 km or 500 meters long. Since we need to convert it to kilometers, let's take 0.5 km as the length of each side.
Therefore, the maximum parking area that can be created is the area of the square:
Area = length * width = (0.5 km)^2 = 0.25 km^2 = 250,000 m^2
So, the maximum parking area that can be created with 2 km of fencing is 250,000 square meters, assuming the rectangle is a square.
To find the maximum parking area that can be created, we need to determine the dimensions of the rectangular area that will maximize the area.
Let's assume the length of the rectangular area (back wall) is x km. Since there are two equal sides, the width of the rectangular area will be (2 km - x km)/2 = (2 - x)/2 km.
The area of a rectangle is given by the formula A = length * width.
So the area equation becomes A = x km * ((2 - x)/2) km = (x(2 - x))/2 km².
To find the maximum area, we can find where the derivative of the area equation is equal to zero. So let's find the derivative:
dA/dx = (2 - 2x)/2 km² = (2 - 2x)/2 km².
Setting dA/dx equal to zero, we have:
(2 - 2x)/2 = 0.
Simplifying the equation, we get:
2 - 2x = 0.
2x = 2.
x = 1 km.
Since x represents the length and we need to calculate the maximum parking area, we substitute x = 1 km back in the area equation:
A = (1(2 - 1))/2 km² = 1 km².
Therefore, the maximum parking area that can be created with 2 km of fencing is 1 km².
To determine the maximum parking area that can be created, we need to find the dimensions of the rectangular area that would use up all 2 km of fencing.
Let's assume the length of the back wall is x. Since there are two equal side lengths for the front and back fences, we can split the remaining length into two equal side lengths, denoted as y.
The length of the back wall will use up x km of fencing.
The two equal side lengths will use up 2y km of fencing (since there are two of them).
So, the total length of fencing used is x + 2y km.
According to the problem, we know that this total length is equal to 2 km:
x + 2y = 2
To maximize the parking area, we need to express the area in terms of a single variable. The area of a rectangle is given by the product of its length and width, which in this case is xy.
To solve for one variable in terms of the other, we can rearrange the equation to solve for x:
x = 2 - 2y
Now we can substitute this expression for x in terms of y into the area formula:
A = xy = (2 - 2y)y = 2y - 2y^2
To find the maximum area, we can use calculus by taking the derivative of the area function with respect to y and setting it equal to zero. However, considering that this problem seems to be solving through basic algebra, we can also solve this equation by completing the square.
To complete the square, we rearrange the equation:
A = -2y^2 + 2y
A = -2(y^2 - y)
Now, we need to add and subtract a constant value inside the parentheses to create a perfect square trinomial:
A = -2(y^2 - y + 1/4 - 1/4)
Using the identity (a - b)^2 = a^2 - 2ab + b^2, we can rewrite the expression as:
A = -2((y - 1/2)^2 - 1/4)
Expanding and simplifying further:
A = -2(y^2 - y + 1/4) + 1/2
A = -2y^2 + 2y - 1/2
Thus, we have expressed the area function as a quadratic equation.
To find the maximum area, we can identify the vertex of this quadratic equation. The vertex has the x-coordinate (in this case, y) and the smallest or largest y-coordinate depending on whether the parabola opens upward or downward.
For a quadratic of the form ax^2 + bx + c, the x-coordinate of the vertex is given by:
x = -b / 2a
In our case, a = -2 and b = 2, so:
y = -2 / [2(-2)] = -1/2
Therefore, the maximum area is achieved when y = -1/2.
Substituting this value into the area function:
A = -2(-1/2)^2 + 2(-1/2) - 1/2
A = -1/2 + 1 - 1/2
A = 0
Hence, the maximum parking area that can be created with 2 km of fencing is 0 km². This means it is not possible to create a parking area with the given constraints.
If the side parallel to the wall has length x meters, and the other side is y, we have
x+2y = 2000
The area a is just
a = xy = (2000-2y)y = 2000y - 2y^2
This is just a parabola, with its vertex (maximum value in this case) at y=500. So, x=1000, and the maximum area is 1000*500 m^2 or 1/2 km^2.
As always happens in this kind of problem, the maximum area for a given perimeter (or, equivalently, the minimum perimeter for a given area) is achieved when the available fence is equally divided between lengths and widths.
In this case, the length is x=1000 and there are two widths of y=500 each, or a total width allocation of 1000.